Challenging +1.8 This is a challenging statics problem requiring multiple equilibrium equations across two connected bodies, geometric reasoning to find angles, and optimization to find the minimum coefficient of friction. It demands systematic application of moments about strategic points, resolution of forces at a smooth joint, and careful handling of limiting friction at two surfaces simultaneously. The multi-body setup with both smooth and rough contacts, plus the need to consider slipping conditions at two locations, places this well above average difficulty for A-level mechanics.
\includegraphics{figure_5}
Two uniform rods \(AB\) and \(BC\) are smoothly jointed at \(B\) and rest in equilibrium with \(C\) on a rough horizontal floor and with \(A\) against a rough vertical wall. The rod \(AB\) is horizontal and the rods are in a vertical plane perpendicular to the wall. The rod \(AB\) has mass \(3m\) and length \(3a\), the rod \(BC\) has mass \(5m\) and length \(5a\), and \(C\) is at a distance \(6a\) from the wall (see diagram). Show that the normal reaction exerted by the floor on the rod \(BC\) at \(C\) has magnitude \(\frac{14}{5}mg\).
[5]
The coefficient of friction at both \(A\) and \(C\) is \(\mu\). Find the least possible value of \(\mu\) for which the rods do not slip at either \(A\) or \(C\).
[7]
Take moments about B for AB: 3aFA = (3a/2)3mg [FA = 3mg/2] M1 A1
Combine to find R C: RC = 8mg – 3mg/2 = 13mg/2 A.G. M1 A1
Substitute for RC in above resln. eqn to find
FA: F A = 8mg – 13mg/2 = 3mg/2 B1
Take moments about B for BC: 4aFC = 3aRC – (3a/2)5mg M1
Substitute for R C to find FC: FC = 3mg A1
Find limiting value µ C for µ at C [or A]
(A.E.F.) : µ C = 6/13 [= 0⋅462 or µ A = 0⋅5] M1 A1
Relate RA , FC by e.g. horizontal resolution: RA = FC [= 3mg] B1
Answer
Marks
Deduce least possible value of µ for system: µ min = max[µ A, µ C] = 0⋅5 B1
5
7
[12]
Question 5:
5 | Resolve forces vertically: RC + FA = 3mg+ 5mg B1
Take moments about B for AB: 3aFA = (3a/2)3mg [FA = 3mg/2] M1 A1
Combine to find R C: RC = 8mg – 3mg/2 = 13mg/2 A.G. M1 A1
Substitute for RC in above resln. eqn to find
FA: F A = 8mg – 13mg/2 = 3mg/2 B1
Take moments about B for BC: 4aFC = 3aRC – (3a/2)5mg M1
Substitute for R C to find FC: FC = 3mg A1
Find limiting value µ C for µ at C [or A]
(A.E.F.) : µ C = 6/13 [= 0⋅462 or µ A = 0⋅5] M1 A1
Relate RA , FC by e.g. horizontal resolution: RA = FC [= 3mg] B1
Deduce least possible value of µ for system: µ min = max[µ A, µ C] = 0⋅5 B1 | 5
7 | [12]
\includegraphics{figure_5}
Two uniform rods $AB$ and $BC$ are smoothly jointed at $B$ and rest in equilibrium with $C$ on a rough horizontal floor and with $A$ against a rough vertical wall. The rod $AB$ is horizontal and the rods are in a vertical plane perpendicular to the wall. The rod $AB$ has mass $3m$ and length $3a$, the rod $BC$ has mass $5m$ and length $5a$, and $C$ is at a distance $6a$ from the wall (see diagram). Show that the normal reaction exerted by the floor on the rod $BC$ at $C$ has magnitude $\frac{14}{5}mg$.
[5]
The coefficient of friction at both $A$ and $C$ is $\mu$. Find the least possible value of $\mu$ for which the rods do not slip at either $A$ or $C$.
[7]
\hfill \mbox{\textit{CAIE FP2 2012 Q5 [12]}}